\newproblem{lay:3_3_32}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.3.32}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $S$ be the tetrahedron in $\mathbb{R}^3$ with vertices at the vectors $\mathbf{0}$, $\mathbf{e}_1$, $\mathbf{e}_2$ and $\mathbf{e}_3$ and
	let $S'$ be the tetrahedron with vertices at vectors $\mathbf{0}$, $\mathbf{v}_1$, $\mathbf{v}_2$ and $\mathbf{v}_3$. See the figure.\\
	\begin{center}
		\includegraphics[scale=0.5]{Tema4/lay_3_3_32.eps}
	\end{center}
	\begin{enumerate}[a.]
		\item Describe a linear transformation that maps $S$ into $S'$.
		\item Find a formula for the volume of the tetrahedron $S'$ using the fact\\
					\begin{center}
						Volume of $S$=$\frac{1}{3}$ Area of the base $\cdot$ Height.
					\end{center}
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item Consider the matrix
					\begin{center}
						$A=\begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \end{pmatrix}$
					\end{center}
					The tetrahedron $S$ is formed by all those points that can be written in the form
					\begin{center}
						$\mathbf{x}=\lambda_0\mathbf{0}+\lambda_1\mathbf{e}_1+\lambda_2\mathbf{e}_2+\lambda_3\mathbf{e}_3$
					\end{center}
					with
					\begin{center}
						$\lambda_0+\lambda_1+\lambda_2+\lambda_3\leq 1$
					\end{center}
					If we consider now $A\mathbf{x}$, we have
					\begin{center}
						$\begin{array}{rcl}A\mathbf{x}&=&A(\lambda_0\mathbf{0}+\lambda_1\mathbf{e}_1+\lambda_2\mathbf{e}_2+\lambda_3\mathbf{e}_3)\\
							 &=&\lambda_0A\mathbf{0}+\lambda_1A\mathbf{e}_1+\lambda_2A\mathbf{e}_2+\lambda_3A\mathbf{e}_3\\
							 &=&\lambda_0\mathbf{0}+\lambda_1\mathbf{v}_1+\lambda_2\mathbf{v}_2+\lambda_3\mathbf{v}_3\end{array}$
					\end{center}
					So this is a point in the tetrahedron $S'$ as required by the problem.
		\item The base of the tetrahedron $S$ is a triangle with vertices $\mathbf{0}$, $\mathbf{e}_1$ and $\mathbf{e}_2$, whose area is
					\begin{center}
						Area triangular base = $\frac{1}{2}$ Base $\cdot$ Height = $\frac{1}{2}1\cdot 1=\frac{1}{2}$.
					\end{center}
					The height of the tetrahedron is the length of $\mathbf{e}_3$, that is, 1. Finally
					\begin{center}
						Volume of $S$=$\frac{1}{3}$ Area of the base $\cdot$ Height = $\frac{1}{3}\frac{1}{2}1=\frac{1}{6}$
					\end{center}
					According to Theorem 5.2 in Chapter 4, the volume of $S'$ is
					\begin{center}
						Volume of $S'$=$|\det\{A\}|$ Volume of $S$=$\frac{1}{6}|\det\{A\}|$
					\end{center}
	\end{enumerate}
}
\useproblem{lay:3_3_32}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
